How to prove $x=120^\circ$ 
Let $ABC$ and $CDE$ be equilateral triangles.
How to prove that $x=120^\circ$?
Thank you.
 A: NOTE: This proof assumes that the segment $AE$ and $BD$ actually intersect at point $F$. Note that the $x=120^{\circ}$ holds when the the lines $AE$ and $BD$, instead of the segments intersect, but you can prove it in a simular fashion.
Let $F$ be the intersection point of $AE$ and $BD$ be $F$. Then from the quadrlaterial $FACD$ we have:
$$\angle AFD + \angle FDC + \angle DCA + \angle CBA = 360^{\circ}$$
$$120^{\circ} + 60^{\circ} - \angle EDF + 60^{\circ} + 60^{\circ} + \angle BCE + 60^{\circ} - \angle BAF = 360^{\circ}$$
$$\angle BCE = \angle BAF + \angle EDF \tag{1}$$
Now for the $\triangle ACE$ we have:
$$\angle EAC + \angle ACE + \angle CEA = 180^{\circ}$$
$$60^{\circ} - \angle BAF + 60^{\circ} + \angle BCE  + \angle CEA = 180^{\circ}$$
Nos using $(1)$ we have:
$$\angle CEA = 60^{\circ} - \angle EDF$$
Now using this we have:
$$\angle FED = 60^{\circ} + \angle CEA = 120^{\circ} - \angle EDF$$
Now using this from the $\triangle FED$ we have:
$$\angle DFE = 180^{\circ} - 120^{\circ} + \angle EDF - \angle EDF$$
$$\angle DFE = 60^{\circ}$$
Now at last:
$$\angle AFD = 180^{\circ} - \angle DFE = 180^{\circ} - 60^{\circ} = 120^{\circ}$$
Q.E.D.
A: Hint:


*

*Observe that $\triangle ACE$ is $\triangle BCD$ rotated by $60^\circ$ degrees.

*The above implies that $|\angle AXB| = 60^\circ$, where $X$ is the intersection of $AE$ and $BD$.


$\hspace{50pt}$
I hope this helps $\ddot\smile$
A: Let intersection of AE and BD be F. G is a point on AE, such that CG is parallel to BD. $\angle CBD = \angle BCG =\beta \Rightarrow \angle GCA =60^\circ - \beta $
Triangles ACE and BCD are equal, therefore $ \angle EAC = \beta $
Therefore $ \angle CGA =120^\circ $    

Triangles ACE and BCD are equal, because two sides and angle between them are equal. 

